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Anonymous
Not applicable

Rename table is not working!!!

Hi guys,

i want to change the name of four tables using this script

For vTableNo = 1 to NoOfTables()-1

let svalue='tab1';

Let vTableName = TableName($(vTableNo)) ;

Rename Table vTableNo to svalue;

next  vTableNo

but nothing happens.

p1.pngswuehl

stalwar1

1 Solution

Accepted Solutions
swuehl
MVP
MVP

I believe RENAME TABLE should use the table name vTableName, not the number vTableNo as source table name.

Using dollar sign expansion might also be required to get the variable values.

Rename Table $(vTableName) to $(svalue);

Then I am not sure what you are trying to achieve here, do you want to rename all tables to the same name?

I don't think that can work (well, haven't tried it).

Or do you want to do something else?

View solution in original post

3 Replies
swuehl
MVP
MVP

I believe RENAME TABLE should use the table name vTableName, not the number vTableNo as source table name.

Using dollar sign expansion might also be required to get the variable values.

Rename Table $(vTableName) to $(svalue);

Then I am not sure what you are trying to achieve here, do you want to rename all tables to the same name?

I don't think that can work (well, haven't tried it).

Or do you want to do something else?

Anonymous
Not applicable
Author

And here is a sample bit of script:

For vTableNo = 0 to NoOfTables() -1

trace vTableNo: $(vTableNo) ;

let vTabNum = $(vTableNo) +1 ;

trace vTabNum: $(vTabNum) ;

let svalue='tab' &  $(vTabNum) ;

trace svalue: $(svalue) ;

Let vTableName = TableName($(vTableNo)) ;

trace vTableName: $(vTableName) ;

Rename Table $(vTableName) to $(svalue) ;

next  vTableNo

The trace command is always useful when debugging thing like this, I have whacked a few in so you can see what it is doing.

Anonymous
Not applicable
Author

thanks very much.

it is working